we don't use 730 at all ?!
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Physics: Post your doubts here!
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How to calculate absolute uncertainties for divisions?
Example, formula = d/v
V is 10 ± 1
D is 13 ± 0.5
Example, formula = d/v
V is 10 ± 1
D is 13 ± 0.5
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this might help.
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Q7)
The relationship needed for this question is the vector relationship F(net) = ma.
This equation tells us that the acceleration of a body is in the same direction as the vector representing the Net Force on that object. The magnitude of this
acceleration is given by F/m. Since the force on the ball is always downwards (the force of gravity, no other force acts on the ball) the acceleration is always downwards too. So, since the question tells us to use the upwards direction as the positive direction, the acceleration is always negative. This acceleration has a magnitude of (mg)/m = 9.81 ms^-2.
Therefore, the only option that has both a constant acceleration in the negative direction and a magnitude of 9.81 ms^-2 is B.
Q9)
We can either eliminate the wrong options, or immediately arrive at the right answer here:
At any point on the trajectory, the object's velocity is along the tangent to that point. So, at the highest point of it's trajectory, the tangent is perfectly horizontal and implies the object has some sort of horizontal velocity.
So, since the object is still moving at the highest point of it's trajectory (it is moving only in the horizontal direction as established above) it has both momentum (in the horizontal direction) and kinetic energy, so the last two options are eliminated. Furthermore, since we have already established that the object is moving in the horizontal direction, B cannot be right. Therefore, A has to be right.
Alternatively, if we apply Newton's Second Law here, F(net) = ma we can see that the only force on the object is in the vertical direction (gravity, exerting a downwards force). Therefore, since there is no force acting on the object that has a component in the horizontal direction, the component of acceleration in the horizontal direction will also be zero, therefore we get A.
I'll finish off the other questions in some time, when i'm free.
Hope this helped!
Good Luck for all your exams!
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can some one tell me how to know that this thing is independent variable and dependent mostly in q1 asked
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plz can some one
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An independent variable is something you physically change; it is something you can control, or something that changes on it's own. It does not depend on any other variables, and it is the value you are supposed to plot on the x-axis during the analysis stage of your experiment.
A dependent variable is a quantity that depends (literally) on the independent variable, or some other property; it usually varies with the independent variable, and it is the value you have no control over; the value of the dependent variable is based on external conditions, usually on the independent variable.
For example:
i) You are measuring the distance a ball travels down an inclined plane (a slope) and the time taken to reach those distances. Your intention is to find how that distance varies with time. Therefore, since time goes independent of any other variable in the experiment, it is the independent variable.
On the same lines of logic, you are measuring the distance traveled, which is the dependent variable because it is dependent on time. It varies with the flow of time, therefore it is the dependent variable.
ii) Suppose you are carrying out an investigation to find out the emf induced in a coil by a magnet falling through it, and want to find out a relation between the emf and the speed of the falling magnet. You have some measure of control over the velocity, and furthermore, the EMF depends on the velocity.
Therefore, since the EMF depends on the velocity (and not the other way around) the velocity is the independent variable and the EMF is the dependent variable.
iii)Imagine you are trying to find the angle a balloon tied to the ground (by a wedge, or any other weight) makes depending on the wind from a fan (This came in the 2012 M/J paper, http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s12_qp_52.pdf)
then clearly you have control over the wind; suppose you are using a fan, you can set the speed, and using a variable resistor you can further vary the speed.
Depending on the speed, the angle made by the balloon will vary.
Since you can control the speed and it is not dependent on any other variable within the bounds of the experiment (the speed depends on the power of the fan, the sensitivity of the setting, etc, but that is not part of the experiment or the relationship you are attempting to find between the needed variables), the wind speed is the independent variable and the angle is the dependent variable.
iv) Suppose you are trying to find out how the peak alternating current varies with the frequency of said current in a circuit with a coil of wire (another paper, this one is http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_s13_qp_52.pdf), there is a clear indicator of the answer in the question itself:
"A student is investigating how the peak alternating current I0 varies with frequency f in a circuit containing a coil of wire."
So he/she is trying to see how I0 changes due to a change in frequency. Therefore, he/she will be changing the frequency and seeing how I0 varies. Therefore, frequency is immediately the independent variable (the variable you can change/vary) and the peak current I0 is the dependent variable.
Hope this helped!
Good Luck for all your exams!
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Q20) Suppose you attach two springs in series (i.e. the ends contact each other) and attach a load to the bottom spring, both springs (assuming their masses are negligible) will display the same extension - the extension of one of these connected springs will be equal to the extension of that spring as if the entire load had been applied on it.
Therefore, the total extension of two springs in series with any load is twice the extension of a single one of those springs (if their spring constants are the same) if the same load had been attached to it.
What you can say here is that since the extension is doubled, the spring-constant for the overall setup is half that of a single spring. For example, suppose you connect two springs with spring constant k in series, the combination will have a spring constant of k/2, i.e. it will be more flexible than a single one of the connected springs. The exact rule here is like connecting resistors in parallel, i.e. 1/k(combination) = 1/k + 1/k = 2/k so k(combination) = k/2.
Suppose you connect the same two springs in parallel, like in arrangement X, then the load is shared among them, so the total extension is half the extension is only one spring had supported the same load. Therefore, if two springs of constant k each are connected in parallel, the combination will have a spring constant of 2k, since they are equivalent to a single, stiff spring.
So, taking this up for the three diagrams X, Y and Z, we find out the equivalent spring constants for each combination (since the load on each is the same, the combination with the highest spring constant will display the least extension):
For X: Since two springs are connected in parallel, their equivalent spring constant is k + k = 2k.
For Y: Since four springs are connected with pairs of two being paired up, and these two parallel branches connected in series to each other, we know that each pair of two has a spring-constant of 2k. Further, 1/k(combination) = 1/(2k) + 1/(2k) = 1/k so that k(combination) = k. (this setup is equivalent to a single spring)
For Z: Three springs are connected; one spring is connected in series to two springs connected in parallel to each other. Therefore, the upper two springs together have a spring constant of 2k. Adding this and the lower single spring in series, we see that 1/k(combination) = 1/(2k) + 1/k = 3/(2k), so that k(combination) = (2k)/3.
Therefore, the smallest extension will be found for setup X (largest spring constant ==> smallest extension) and the next smallest extension will be found for Y. Therefore, the largest extension will be found for setup Z. So the answer is A.
Q22)
Both wires obey Hooke's Law - this is important, since it means they also obey the formula of Young's Modulus, which we can apply as follows:
Since both wires are made of the same material, they have the same value of Young's Modulus (Y).
Since both wires have the same extension, they have the same value of e in the formula Y = Fl/Ae (Or Fl/Ax , whichever one you prefer).
Therefore, we can equate their extensions and cancel out the value of Y or equate the values of Y and cancel out the value of e. Let's do the former.
So, for P: (Tension in P) * l / AY = extension
And for Q: (Tension in Q) * (2l) / (A/2)Y = extension
So (Tension in P) * l / AY = (Tension in Q) *2l/ (A/2)Y
Cancelling out values on both sides,
(Tension in P) * 1 = 4 * (Tension in Q)
So that (Tension in P)/(Tension in Q) = 4/1 = D.
Q24) This concerns values you need to keep in memory unless they are on the data sheets, i'm not sure. Either ways, visible light falls in the range of
400 nm< λ < 700 nm. We can take any value in this range to represent out wavelength. Suppose we take 600 nm.
Then, the number of wavelengths in 1 meter = 1 / (600 * 10^-9) = 1.6667 * 10^6. So our answer is closest to B. Therefore, that's our answer.
I'll post the rest after some more time, have some work I need to finish off
Hope this helped!
Good Luck for all of your exams!
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From the diagram, we can see that resistors A and B are connected to each other in series. Therefore, their total resistance R(A+B) is equal to R + R = 2R.
Furthermore, together they are connected in parallel to D, so 1/R(A+B+D) = 1/(2R) + 1/(R) = 1/(2R) + 2/(2R) = 3/(2R)
Therefore, R(A+B+D) = 2R/3.
This resistance, R(A+B+D) is connected in a series arrangement to resistor C, so their total resistance R(A+B+C+D) = (2R)/3 + R = (2R + 3R)/3 = 5R/3
This overall is connected in parallel to resistor E, so the resistance R(A+B+C+D+E) is given by
1/R(A+B+C+D+E) = 1/R(A+B+C+D) + 1/R(E) = 3/(5R) + 1/R = 3/(5R) + 5/(5R) = 8/(5R)
Therefore, R(A+B+C+D+E) = 5R/8 ==> C.
Hope this helped!
Good Luck for all your exams!
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(D) Workdone = force(F) x displacement(x) hence, x = W/f
Here W = K.E of electron = 1/2 x m x v^2 and F = qE = eE where e is cahrge of electron and E is electric field strength.
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Thanks but in q22 How to cancel values ? ( I am very very weak )Q20) Suppose you attach two springs in series (i.e. the ends contact each other) and attach a load to the bottom spring, both springs (assuming their masses are negligible) will display the same extension - the extension of one of these connected springs will be equal to the extension of that spring as if the entire load had been applied on it.
Therefore, the total extension of two springs in series with any load is twice the extension of a single one of those springs (if their spring constants are the same) if the same load had been attached to it.
What you can say here is that since the extension is doubled, the spring-constant for the overall setup is half that of a single spring. For example, suppose you connect two springs with spring constant k in series, the combination will have a spring constant of k/2, i.e. it will be more flexible than a single one of the connected springs. The exact rule here is like connecting resistors in parallel, i.e. 1/k(combination) = 1/k + 1/k = 2/k so k(combination) = k/2.
Suppose you connect the same two springs in parallel, like in arrangement X, then the load is shared among them, so the total extension is half the extension is only one spring had supported the same load. Therefore, if two springs of constant k each are connected in parallel, the combination will have a spring constant of 2k, since they are equivalent to a single, stiff spring.
So, taking this up for the three diagrams X, Y and Z, we find out the equivalent spring constants for each combination (since the load on each is the same, the combination with the highest spring constant will display the least extension):
For X: Since two springs are connected in parallel, their equivalent spring constant is k + k = 2k.
For Y: Since four springs are connected with pairs of two being paired up, and these two parallel branches connected in series to each other, we know that each pair of two has a spring-constant of 2k. Further, 1/k(combination) = 1/(2k) + 1/(2k) = 1/k so that k(combination) = k. (this setup is equivalent to a single spring)
For Z: Three springs are connected; one spring is connected in series to two springs connected in parallel to each other. Therefore, the upper two springs together have a spring constant of 2k. Adding this and the lower single spring in series, we see that 1/k(combination) = 1/(2k) + 1/k = 3/(2k), so that k(combination) = (2k)/3.
Therefore, the smallest extension will be found for setup X (largest spring constant ==> smallest extension) and the next smallest extension will be found for Y. Therefore, the largest extension will be found for setup Z. So the answer is A.
Q22)
Both wires obey Hooke's Law - this is important, since it means they also obey the formula of Young's Modulus, which we can apply as follows:
Since both wires are made of the same material, they have the same value of Young's Modulus (Y).
Since both wires have the same extension, they have the same value of e in the formula Y = Fl/Ae (Or Fl/Ax , whichever one you prefer).
Therefore, we can equate their extensions and cancel out the value of Y or equate the values of Y and cancel out the value of e. Let's do the former.
So, for P: (Tension in P) * l / AY = extension
And for Q: (Tension in Q) * (2l) / (A/2)Y = extension
So (Tension in P) * l / AY = (Tension in Q) *2l/ (A/2)Y
Cancelling out values on both sides,
(Tension in P) * 1 = 4 * (Tension in Q)
So that (Tension in P)/(Tension in Q) = 4/1 = D.
Q24) This concerns values you need to keep in memory unless they are on the data sheets, i'm not sure. Either ways, visible light falls in the range of
400 nm< λ < 700 nm. We can take any value in this range to represent out wavelength. Suppose we take 600 nm.
Then, the number of wavelengths in 1 meter = 1 / (600 * 10^-9) = 1.6667 * 10^6. So our answer is closest to B. Therefore, that's our answer.
I'll post the rest after some more time, have some work I need to finish off
Hope this helped!
Good Luck for all of your exams!
In q24 why you took 600 nm ? Y not 400nm or any other value ?
Thanks again, And please post the rest answers.
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Guyssss plz heeeelpp , how to find v and the uncertainty of v^2 in Q2 ????
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_53.pdf
http://papers.xtremepapers.com/CIE/...nd AS Level/Physics (9702)/9702_w11_qp_53.pdf
Check this site guys! Really helpful for finding uncertainties.
http://spiff.rit.edu/classes/phys273/uncert/uncert.html
Credits: sagar65265
http://spiff.rit.edu/classes/phys273/uncert/uncert.html
Credits: sagar65265